Question

The value of integral $\underset{C}{\oint }\frac{si{n}^{2}z}{{(z-\frac{\pi }{2})}^3}dz$, where C is the |z| =1 , will be

• A. $2\pi i$
• B. $\pi i$
• C. 0
• D. $-\pi i$

Answer 1

The correct option is C 0

(c)

The poles at $z=\frac{\pi }{2}=1.57$ lies outside the circle |Z| =1.

$So,\underset{C}{\oint }\frac{si{n}^{2}z}{{(z-\frac{\pi }{2})}^3}dz=0$